You have found the following ages (in years) of 4 zebras. Those zebras were randomly selected from the 49 zebras at your local zoo: $ 7,\enspace 20,\enspace 1,\enspace 2$ Based on your sample, what is the average age of the zebras? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we only have data for a small sample of the 49 zebras, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $4$ samples and divide by $4$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\overline{x}} = \dfrac{7 + 20 + 1 + 2}{{4}} = {7.5\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {0.25} + {156.25} + {42.25} + {30.25}} {{4 - 1}} $ {s^2} = \dfrac{{229}}{{3}} = {76.33\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{76.33\text{ years}^2}} = {8.7\text{ years}} $ We can estimate that the average zebra at the zoo is 7.5 years old. There is also a standard deviation of 8.7 years.